Prove that $\sqrt{3}+\sqrt{5}$ is an irrational number.

To do:

We have to prove that $\sqrt{3}+\sqrt{5}$ is an irrational number.

Solution:

Let $x=\sqrt{3}+\sqrt{5}$ be a rational number.

Squaring on both sides, we get,

$x^{2}=(\sqrt{3}+\sqrt{5})^{2}$

$x^{2}=3+5+2 \times \sqrt{3} \times \sqrt{5}$

$x^{2}=8+2 \sqrt{15}$

$x^{2}-8=2 \sqrt{15}$

$\Rightarrow \frac{x^{2}-8}{2}=\sqrt{15}$

$x=\sqrt{3}+\sqrt{5}$ is a rational number.

This implies,

$\frac{x^{2}-8}{2}$ should be a rational number.

$\Rightarrow \sqrt{15}$ is a rational number.
But it is not possible as $\sqrt{15}$ is an irrational number.

Therefore, our assumption that $x=\sqrt{3}+\sqrt{5}$ is a rational number is false.

Hence, $\sqrt{3}+\sqrt{5}$ is an irrational number. 

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Updated on: 10-Oct-2022

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